The square of the distance of the point of intersection of the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}$ and the plane $2x-y+z=6$ from the point $(-1,-1,2)$ is .... .

Let $\vec{r} \cdot (3\hat{i} - 2\hat{j} + 7\hat{k}) = 32$ be the equation of a plane. If the line with direction ratios $(5, b, 3)$ is parallel to the plane, then the value of $b$ is:

The lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-K}$ and $\frac{x-1}{K}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar if

The position vector of the point of intersection of the line $\bar{r}=(2 \hat{i}+\hat{j}-4 \hat{k})+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$ and the $XOY$-plane is:

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect,then $k$ is equal to:

If for $a > 0,$ the feet of perpendiculars from the points $A(a, -2a, 3)$ and $B(0, 4, 5)$ on the plane $lx + my + nz = 0$ are points $C(0, -a, -1)$ and $D$ respectively,then the length of line segment $CD$ is equal to